Class 12 Maths | Theory of Equations Exercise 3.1 | Q.6 to Q.10 | TN State Board #mathematics

Added: 2026-05-19

[00:00:02]Dear students, most welcome to our channel.

[00:00:07]Chapter 3, theory of equations.

[00:00:13]Exercise 3.1 question number.

[00:00:18]So continuation question number six solve the equation pol if it is given that two of its roots are in the ratio 3 is to 2.

[00:00:41]So fundamental of algebra for examp 15 15 So let 3 alpha and 2 alpha and beta are the roots of above equation.

[00:01:36]So sum of sum of roots taken at two two at a time. Next product of roots.

[00:01:58]Sum of coefficient 14 answer 15. Next sum of coefficient of evenus 24. So answer 24 - 9 15. So in the particular Sum of odd degree coefficient.

[00:02:57]This is equal to sum of even degree coefficient.

[00:03:12]So equal. Therefore, minus1 must be the root of the given cubic polomial equation. xq - 9 x² + 14x + 24 which is equal to 0.

[00:03:42]Just descending order 3 2 1 x^ 1 - 9 + 14 + 24us - 10.

[00:04:16]Next minus into plus 1 into 10 - 24us.

[00:04:29]So evidence equation x² - 10 x + 24= z remaining two roots.

[00:04:57]So and remaining two roots. So simple 24us 24us.

[00:05:13]So 24 into 4 24 plus.

[00:05:23]So give both minus signus plus x - 6 x - 4 which is = 0 x - 6 = 0 and then x - 4 = x = 6 = 4. So finally roots are first minus one.

[00:05:57]Next 6 4 6 4 3.

[00:06:10]So final answer.

[00:06:20]Okay. Next. Next question number seven.

[00:06:27]Exam oriented sum.

[00:06:29]If alpha, beta, gamma are the roots of the polomial equation that is cubic polomial equation.

[00:06:40]beta roots. So next find the value of summation alpha / beta into gamma in the terms of coefficient.

[00:06:57]So given alpha, beta and gamma are the roots of the given cubic polomial equation general equation a x² b x² + c x + d which is equal to zero. So for sum of roots summation of alpha meaning alpha plus beta plus gamma. So sum of roots minus b by a coefficient.

[00:07:50]Next sum of roots taken two at a time.

[00:07:56]Alpha beta expand alpha beta plus beta gamma plus gamma alpha this is equal minus b by plus c by a next product of roots that is alpha beta gamma this is equal to minus plus next minus min - b by a So summation of alpha / beta into gamma.

[00:08:31]Therefore, summation of alpha / beta into gamma exp.

[00:08:50]Next.

[00:08:59]Next.

[00:09:11]answer.

[00:09:16]This is equal simply take LCM alpha beta gamma denominator simply alpha square.

[00:09:54]This is equation number one summation of beta minus t by a That is alpha squared plus beta squ + gamma square + beta + gamma square a² + B² + C square A + B + C square for + 2 into A B + B C plus CA alpha beta plus beta gamma plus gamma alpha beta + gamma b so - b by a whole square.

[00:11:11]The answer we know that is alpha square beta square gamma square + 2 into alpha beta plus beta gamma plus gamma c that is c by a this is b² by a square by a this is equal to summation of ala square.

[00:11:44]So final LCM a square a square B square answer. So equation number one suppose that is summation of alpha / beta into gamma.

[00:12:30]So alpha square + beta² + gamma square divided by alpha beta gamma sol. So value so denominator product of roots d by a numerator two terms b² - 2 c a / a² b² - 2 c a / a² into - a / E so A² A cancel. So final answer b² - 2 a c / a into d a minus so minus a minus n denominator 2 a minus b square by a so answer.

[00:13:47]So next question number eight alpha, beta, gamma and delta are the roots of the polomial equation. Fourth degree polation with integer coefficient that is very very important number find a quot equation with integer coefficients whose roots are so either rooted First given answer given alpha, beta, gamma and delta are the roots of the given equation.

[00:14:58]More General equation 2 x^ 4 + 5 xq - 7 x² + 8 which= 0.

[00:15:17]Therefore sum of roots alpha + beta + gamma + delta this is equal to usual formula -2 next.

[00:15:46]So product of roots alpha beta gamma delta this is equal x² the product of roots the degree -1 Even number.

[00:16:38]So product of roots.

[00:16:43]So 8 / 2. So let als form of a quadratic equation.

[00:17:07]General form of a quadratic equation.

[00:17:13]So X² of roots into X of roots of roots P + Q into X. Next product of roots P into Q. This is equal to Z form.

[00:17:46]P + Q P + Q. So this is equal value - 4 - 5 by 2 + 4. So + 8 - 5. So answer 3 by 2 product of roots P into Q - 5 by 2 into 4. So this is equal 2.

[00:18:16]Sous. So just answer. So one implies. So required quadratic equation x² - s of root into x that is 3x 2 into x. Next product of roots simply - 10 which is equal to z. So next we need with integer coefficient.

[00:18:59]both side multip 2x² - 3x - 20 which is equal to 0.

[00:19:24]So next question question number n. So important if the equations x² + px + q =0 and x² + p - x + qd =0 have a common root show that it must be equal to Answer root value simp.

[00:20:14]Sorry.

[00:20:18]assumption.

[00:20:21]So assumption let alpha x² + px + q = 0. Next equation x² + p - x + q dash which is = 0 given given one root is common of these two equations.

[00:21:04]Let alpha be the common root of equation.

[00:21:16]Let alpha be the common root of equation one and two.

[00:21:31]equation exist square plus P into alpha + Q which is equal to zero. Next equation number two square + pdash into alpha + q ddash which is equal to z cross multiplication But easiest method crossic Q dash next P dash square. Next term four terms multiplication alpha square only the four terms.

[00:23:26]So first cross multiplication cross multiplication P into Q dash P Q dash P into Q plus alternate signus P into Q this is equal next 1 into Q Q Next.

[00:24:04]This is equal. Next final four terms.

[00:24:16]So 1 P.

[00:24:23]So that is final answer alpha square divided by p Q dash minus pdash q this is equal to 1 / pdash minus B.

[00:25:00]So first two terms.

[00:25:20]So Second term. So alpha / q - q dash.

[00:25:29]Next part next second and third alpha / q - q dash which is equal to 1 / pus square.

[00:25:51]Therefore, alpha which is equal to in the denominator right P QA minus P dash Q divided by Q minus Q dash. So the first value next that is alpha= q - q dash / pus ala = q - q dash divid pd minus p So cross multiplication elimination question number three or four steps concept.

[00:27:02]A 12 m tree tall tree was broken into two parts. It was found that the height of the pot which was left standing was the cube roots of the length of the pot was cut away. This is very very important word standing that is the cube roots cube roots of the length of the part that was cut away.

[00:27:35]Formulate this into a mathematical problem to find the height of the part which was cut away.

[00:27:48]data 12 m tall tree.

[00:28:20]So standing in the mathematical formula formula.

[00:28:50]So maxable XY Zeose 2 m.

[00:29:25]So that is 12 - X relationship X= roots of 12 - X cube roots of X cube roots That is x + 12 = 2 remaining part.

[00:30:40]So next thank you. Thank you for watching.